Question

A non-uniform rod is balanced as shown in the diagram. Given that the center of mass of the rod is 14.0 cm from the end marked A, what is the mass of the rod?

a) 40g b) 100g c) 1000g d) 1140g

Answer 1

Given:

A non-uniform rod is balanced as shown in the diagram. Given that the center of mass of the rod is 14.0 cm from the end marked A.

To find:

Let the mass be m. Now, the gravitational force will act on the CENTRE OF MASS.

Now, since the rod is in rotational equilibrium, the net torque will be zero.

Considering the axis of rotation at the ∆:

$\therefore \: \sum( \tau) = 0$

$\implies \: mg(30 - 14) - \dfrac{80}{1000} g(20) = 0$

$\implies \: mg(16) - \dfrac{80}{1000} g(20) = 0$

$\implies \: m(16) - \dfrac{80}{1000} (20) = 0$

$\implies \: m(16) - \dfrac{8}{10} (2) = 0$

$\implies \: m(16) = \dfrac{8}{10} (2)$

$\implies \: m(2) = \dfrac{1}{10} (2)$

$\implies \: m = \dfrac{1}{10} \: kg$

$\implies \: m = \dfrac{1}{10} \times 1000 \: gram$

$\implies \: m = 100 \: gram$

So, mass of rod is 100 grams.